SQL Challenges-1: Find the highest salaried employees for each designation

Last update on August 19 2022 21:51:03 (UTC/GMT +8 hours)

SQL Challenges-1: Exercise-67 with Solution

From the following table write a query in SQL to find the highest salaried employees for each designation. Return job_id, name and the salary earned.

Table: employees

Structure:

Field	Type	Null	Key
employee_id	int	NO	PRI
emp_name	varchar(25)	YES
hire_date	date	YES
job_id	varchar(25)	YES
salary	decimal(10,2)	YES
manager_id	int	YES
department_id	int	YES

Data:

employee_id	emp_name	hire_date	job_id	salary	manager_id	department_id
100	Steven	1987-06-17	AD_PRES	24000.00	0	90
101	Neena	1987-06-18	AD_VP	17000.00	100	90
102	Lex	1987-06-19	AD_VP	17000.00	100	90
103	Alexander	1987-06-20	IT_PROG	9000.00	102	60
104	Bruce	1987-06-21	IT_PROG	6000.00	103	60
105	David	1987-06-22	IT_PROG	4800.00	103	60
106	Valli	1987-06-23	IT_PROG	4800.00	103	60
107	Diana	1987-06-24	IT_PROG	4200.00	103	60
108	Nancy	1987-06-25	FI_MGR	12000.00	101	100
109	Daniel	1987-06-26	FI_ACCOUNT	9000.00	108	100
110	John	1987-06-27	FI_ACCOUNT	8200.00	108	100
111	Ismael	1987-06-28	FI_ACCOUNT	7700.00	108	100
112	Jose Manuel	1987-06-29	FI_ACCOUNT	7800.00	108	100
113	Luis	1987-06-30	FI_ACCOUNT	6900.00	108	100
114	Den	1987-07-01	PU_MAN	11000.00	100	30
115	Alexander	1987-07-02	PU_CLERK	3100.00	114	30
116	Shelli	1987-07-03	PU_CLERK	2900.00	114	30
117	Sigal	1987-07-04	PU_CLERK	2800.00	114	30
133	Jason	1987-07-20	ST_CLERK	3300.00	122	50
134	Michael	1987-07-21	ST_CLERK	2900.00	122	50
135	Ki	1987-07-22	ST_CLERK	2400.00	122	50
136	Hazel	1987-07-23	ST_CLERK	2200.00	122	50
137	Renske	1987-07-24	ST_CLERK	3600.00	123	50
138	Stephen	1987-07-25	ST_CLERK	3200.00	123	50
139	John	1987-07-26	ST_CLERK	2700.00	123	50

Sample Solution:

SQL Code(MySQL):

create table employees (
employee_id integer(4) not null unique,
emp_name varchar(25),
hire_date date,
job_id varchar(25),
salary decimal(10,2),
manager_id integer(4),
department_id integer(4)
);



insert into employees values( 100,'Steven     ','1987-06-17','AD_PRES   ',24000.00,         0,   90);
insert into employees values( 101,'Neena      ','1987-06-18','AD_VP     ',17000.00,       100,   90);
insert into employees values( 102,'Lex        ','1987-06-19','AD_VP     ',17000.00,       100,   90);
insert into employees values( 103,'Alexander  ','1987-06-20','IT_PROG   ', 9000.00,       102,   60);
insert into employees values( 104,'Bruce      ','1987-06-21','IT_PROG   ', 6000.00,       103,   60);
insert into employees values( 105,'David      ','1987-06-22','IT_PROG   ', 4800.00,       103,   60);
insert into employees values( 106,'Valli      ','1987-06-23','IT_PROG   ', 4800.00,       103,   60);
insert into employees values( 107,'Diana      ','1987-06-24','IT_PROG   ', 4200.00,       103,   60);
insert into employees values( 114,'Den        ','1987-07-01','PU_MAN    ',11000.00,       100,   30);
insert into employees values( 115,'Alexander  ','1987-07-02','PU_CLERK  ', 3100.00,       114,   30);
insert into employees values( 116,'Shelli     ','1987-07-03','PU_CLERK  ', 2900.00,       114,   30);
insert into employees values( 117,'Sigal      ','1987-07-04','PU_CLERK  ', 2800.00,       114,   30);
insert into employees values( 108,'Nancy      ','1987-06-25','FI_MGR    ',12000.00,       101,  100);
insert into employees values( 109,'Daniel     ','1987-06-26','FI_ACCOUNT', 9000.00,       108,  100);
insert into employees values( 110,'John       ','1987-06-27','FI_ACCOUNT', 8200.00,       108,  100);
insert into employees values( 111,'Ismael     ','1987-06-28','FI_ACCOUNT', 7700.00,       108,  100);
insert into employees values( 112,'Jose Manuel','1987-06-29','FI_ACCOUNT', 7800.00,       108,  100);
insert into employees values( 113,'Luis       ','1987-06-30','FI_ACCOUNT', 6900.00,       108,  100);
insert into employees values( 133,'Jason      ','1987-07-20','ST_CLERK  ', 3300.00,       122,   50);
insert into employees values( 134,'Michael    ','1987-07-21','ST_CLERK  ', 2900.00,       122,   50);
insert into employees values( 135,'Ki         ','1987-07-22','ST_CLERK  ', 2400.00,       122,   50);
insert into employees values( 136,'Hazel      ','1987-07-23','ST_CLERK  ', 2200.00,       122,   50);
insert into employees values( 137,'Renske     ','1987-07-24','ST_CLERK  ', 3600.00,       123,   50);
insert into employees values( 138,'Stephen    ','1987-07-25','ST_CLERK  ', 3200.00,       123,   50);
insert into employees values( 139,'John       ','1987-07-26','ST_CLERK  ', 2700.00,       123,   50);




SELECT a.job_id, emp_name AS Name, salary
FROM employees a
JOIN
(SELECT job_id, MAX(salary) As max_salary
FROM employees
GROUP BY 1 ) b
ON a.job_id=b.job_id AND a.salary=b.max_salary;

Sample Output:

job_id    |Name       |salary  |
----------+-----------+--------+
AD_PRES   |Steven     |24000.00|
AD_VP     |Neena      |17000.00|
AD_VP     |Lex        |17000.00|
IT_PROG   |Alexander  | 9000.00|
FI_MGR    |Nancy      |12000.00|
FI_ACCOUNT|Daniel     | 9000.00|
PU_MAN    |Den        |11000.00|
PU_CLERK  |Alexander  | 3100.00|
ST_CLERK  |Renske     | 3600.00|

SQL Code Editor:

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